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25 Nov 2021, 10:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:29) correct
63%
(02:12)
wrong
based on 67
sessions
History
Date
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Not Attempted Yet
A merchant has 3 kinds of wine. 403 gallons of first kind, 527 gallons of second kind & 589 gallons of third kind. Find the least number of casks of equal size in which the wine can be stored without mixing.
(A) 49
(B) 31
(C) 19
(D) 17
(E) 13
(A) 49
(B) 31
(C) 19
(D) 17
(E) 13
PyjamaScientist
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Joined: 25 Oct 2020
Last visit: 04 Apr 2026
Posts: 1,125
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Schools: Ross '25 (M$)
GMAT 1: 740 Q49 V42 (Online)
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25 Nov 2021, 10:36
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For highest capacity container we need the H.C.F. of all amounts of the wine.
H.C.F. of 403, 527 and 589 = 31
Now the number of full casks required = \(\frac{403+527+589}{31}\) = 49 (A)
H.C.F. of 403, 527 and 589 = 31
Now the number of full casks required = \(\frac{403+527+589}{31}\) = 49 (A)
25 Nov 2021, 10:58
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Technically the answer is 3, because you can just use three 600-gallon casks (or any other size 589 gallons or greater). If the casks need to be filled completely, then pyjamascientist's solution is perfect.
